Problem: Solve for $x$ : $ 3|x + 1| - 9 = -6|x + 1| + 3 $
Explanation: Add $ {6|x + 1|} $ to both sides: $ \begin{eqnarray} 3|x + 1| - 9 &=& -6|x + 1| + 3 \\ \\ { + 6|x + 1|} && { + 6|x + 1|} \\ \\ 9|x + 1| - 9 &=& 3 \end{eqnarray} $ Add ${9}$ to both sides: $ \begin{eqnarray} 9|x + 1| - 9 &=& 3 \\ \\ { + 9} &=& { + 9} \\ \\ 9|x + 1| &=& 12 \end{eqnarray} $ Divide both sides by ${9}$ $ \dfrac{9|x + 1|} {{9}} = \dfrac{12} {{9}} $ Simplify: $ |x + 1| = \dfrac{4}{3}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 1 = -\dfrac{4}{3} $ or $ x + 1 = \dfrac{4}{3} $ Solve for the solution where $x + 1$ is negative: $ x + 1 = -\dfrac{4}{3} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& -\dfrac{4}{3} \\ \\ {- 1} && {- 1} \\ \\ x &=& -\dfrac{4}{3} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $3$ $ x = - \dfrac{4}{3} {- \dfrac{3}{3}} $ $ x = -\dfrac{7}{3} $ Then calculate the solution where $x + 1$ is positive: $ x + 1 = \dfrac{4}{3} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& \dfrac{4}{3} \\ \\ {- 1} && {- 1} \\ \\ x &=& \dfrac{4}{3} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $3$ $ x = \dfrac{4}{3} {- \dfrac{3}{3}} $ $ x = \dfrac{1}{3} $ Thus, the correct answer is $x = -\dfrac{7}{3} $ or $x = \dfrac{1}{3} $.